Question: The sum of the first 2011 terms of a geometric sequence is 200. The sum of the first 4022 terms is 380. Find the sum of the first 6033 terms.
Let the first term be $a,$ and let the common ratio be $r.$  Then
\[a + ar + ar^2 + \dots + ar^{2010} = 200\]and
\[a + ar + ar^2 + \dots + ar^{4021} = 380.\]Subtracting these equations, we get
\[ar^{2011} + ar^{2012} + \dots + ar^{4021} = 180.\]Then
\[r^{2011} (a + ar + \dots + ar^{2010}) = 180,\]so
\[r^{2011} = \frac{180}{200} = \frac{9}{10}.\]Then the sum of the first 6033 terms is
\begin{align*}
a + ar + ar^2 + \dots + ar^{6032} &= (a + ar + ar^2 + \dots + ar^{4021}) + (ar^{4022} + ar^{4023} + \dots + ar^{6032}) \\
&= 380 + r^{4022} (a + ar + \dots + ar^{2010}) \\
&= 380 + \left( \frac{9}{10} \right)^2 \cdot 200 \\
&= \boxed{542}.
\end{align*}